$A=\left[\begin{array}{rr}21 & -12 & 18 & 6 \\-8 & 3 & 4 & -1 \\2 &2 &-8 & 4 \\1 &9 &-15 & -24 \\2 &2 &1 & 4\end{array}\right]$ $A_{5,1}=$
Solution: Background An $m\times n$ matrix has $m$ rows and $n$ columns. $A=\left[\begin{array}{rr}A_{1,1} & \cdots & A_{1,n} \\\\\vdots \ & \ddots & \vdots \\\\A_{m,1} &\cdots &A_{m,n}\end{array}\right]$ Therefore, the entry $A_{{c},{d}}$ is located on row ${c}$ and column ${d}$. Finding $A_{5,1}$ $A_{{5},{1}}$ is located on row ${5}$ of $A$ : $\left[\begin{array}{rr}21 & -12 & 18 & 6 \\-8 & 3 & 4 & -1 \\2 &2 &-8 & 4 \\1 &9 &-15 & -24 \\ {2} & {2} & {1} & {4}\end{array}\right]$ $A_{{5},{1}}$ is also located on column ${1}$ of $A$. $\left[\begin{array}{rr}{21} & -12 & 18 & 6 \\{-8} & 3 & 4 & -1 \\{2} &2 &-8 & 4 \\{1} &9 &-15 & -24 \\ {\text2} & {2} & {1} & {4}\end{array}\right]$ Therefore, $A_{{5},{1}}={2}$. Summary $A_{5,1}=2$